Download Special Theory of Relativity

Special Theory of Relativity
“I had a better grasp of things when physics dealt
mostly with falling bodies.”
The Newtonian Electron
• Newtonian Theory (everything we have done
so far in class) can be tested at high speeds
by accelerating electrons or other charged
particles through a potential difference.
• However, experiments have shown, that no
matter the size of the accelerating voltage,
the speed of the electron (or any other
particle with mass) will always be less then
the speed of light.
• This means nothing with mass can go
faster then 3.0 x 108 m/s
Universal
Speed
Limit
3.0x108m/s
Newton vs. Einstein
• Since Newtonian theory no longer worked at
high speed, another theory was needed.
• This is where Einstein stepped in.
• In 1915 Einstein published his general theory
of relativity.
– Even though this theory is what Einstein is mostly
known for, it is not what won him a Nobel Prize
– His Nobel Prize was for his explanation of the
Photoelectric Effect in 1921
So if Newtonian Physics is
wrong….
• As long as an object’s speed is much
less then the speed of light, Newtonian
Physics works wonderfully.
• However, if an object’s speed starts to
approach the speed of light some
interesting things occur.
It is all Relative
• According to the Special Theory of
Relativity, two observers moving relative
to each other, will measure different
outcomes for the same event
• This makes it necessary to choose a
frame of reference
Inertial Frames of Reference
are:
• Any reference frame in which
Newtonian Physics is valid.
• Any reference frame in which objects
that experience no forces, move in
straight lines at a constant speed (or not
at all).
• We will be working with these types of
reference frames.
Example: Two students are playing baseball on a train
moving at 100 mi/hr. The pitcher throws the ball at 50 mi/hr.
According to a stationary observer, how fast is the ball going?
50 mi/hr
100 mi/hr
Baseball
speed =
150 mi/hr
The batter hits it back at 50 mi/hr.
50 mi/hr
100 mi/hr
Baseball
speed =
50 mi/hr
What if instead of a baseball,
it was a light pulse?
But remember, nothing can go faster then
3.0x108 m/s (including light itself).
Therefore, there must be a problem with the
classical addition law for velocities.
And that is where Einstein’s Special Theory of
Relativity comes in.
Relativities Two Postulates
1. The principle of Relativity: All the laws of
physics are the same in all inertial reference
frames.
2. The constancy of the speed of light: The
speed of light in a vacuum has the same
value (c = 2.997 924 58x108m/s, rounded to
3.0 in this class) in all inertial reference
frames, regardless of the velocity of the
observer or the velocity of the source
emitting the light.
Relativistic Addition of Velocity
• The problem: A motorcycle moving at
0.80c with respect to a stationary
observer. Then the rider throws a ball
forward at 0.70c relative to himself.
• According to Newton (and classical
mechanics), how fast is the ball going?
The solution:
Velocity of object
relative to
a stationary
person
vom  vms
vos 
vom vms
1 2
c
Velocity of object
relative to
a moving
person

Velocity of mover
relative to
a stationary
person
Object – what is being “launched”
Mover – what does the
“launching”
Stander – Usually the Earth
Example: A motorcycle moving at 0.80c with respect to
a stationary observer. Then the rider throws a ball
forward at 0.70c relative to himself. How fast is the ball
going relative to a person standing still?
• Vos = ?
• Vms = 0.80c
• Vom = 0.70c
vom + vms
vos =
vomvms
1+ 2
c
0.70c  0.80c
vos 
 0.96c
(0.70c)(0.80c)
1
2
c
Example 2: A spaceship leaves Earth traveling at 0.71c relative to
Earth. A second mini spaceship leaves the first at a speed of 0.87c
with respect to the first. Calculate the speed of the second ship with
respect to Earth if it is launched in the opposite direction, back
towards Earth.
• Vos = ?
• Vms = 0.71c
• Vom = - 0.87c
vom  vms
vos 
vom vms
1 2
c
 0.87c  0.71c
vos 
 0.42c
(0.87c)(0.71c)
1
2
c

Effects of Relativity
• Length contraction - Moving rulers are short
• Time dilation - Moving clocks run slow
Length Contraction
• When viewed by an outside observer, moving
objects appear to contract along the direction
of motion.
• For everyday speeds, the amount of
contraction is too small to be measured.
• For relativistic speeds, the contraction is
noticeable.
– A meter stick whizzing past you on a spaceship
moving at 87% the speed of light (0.87c) would
appear to be only 0.5 m long.
What would a baseball thrown at
relativistic speeds look like to a fan sitting
in the stands?
• It would contract along the direction of motion.
At rest
0.50c
.95c
Calculating the length
contraction:
v 2 
L  Ls 1  2 
L = moving length
c 
Ls = stationary length (length at rest)
v = velocity of the object
c = speed of light

Example: A meter stick flies past you at 99.5%
the speed of light. What is it’s apparent length?
v 2 
L  Ls 1  2 
c 

.9952 c 2 
L 1 1  2  1 0.990
 c

L = 0.0999 m or 9.99 cm

How do people on spaceships
view their meter sticks?
1. They are smaller then usual
2. They are the same
3. They are larger than life
You are packing for a trip to another star, and
on your journey you will be traveling at a speed
of 0.99c. Can you sleep in a smaller cabin then
usual, because you will be shorter when you lie
down? Explain.
Time Dilation
• Pretend you are in a spaceship at rest
in Ms. Keeler’s class. The clock on the
wall reads 12-noon.
• To say it reads “12 noon” is to say that
light reflects from the clock and carries
the information “12 noon” to you in the
direction of sight.
• If you suddenly move your head to the
side, the light would miss your eye and
continue out into space where another
observer might see it.
• The observer in space would then later
say “Oh it is 12 noon on Earth right
now”
• But from your point of view, it isn’t.
• Now suppose your spaceship is moving
as fast as the speed of light (just
pretend!).
• You would be keeping up with the signal
saying “12 noon”
• To you on the spaceship, time at home
would appear frozen!
• This is in essence, time dilation:
• Clocks that are moving, run slow.
• Note: When a clock is “running slow” it shows
a smaller time than it “should”
Consider a light clock
• Light comes out of the
source, hits a mirror
and bounces back
into a detector.
• Based on how far it
traveled and the
speed of light, we
could calculate the
time it took.
Now, put the time clock on a spaceship…
Since the speed of
light is the same
for everyone, time
must be running
slow for the
astronaut.
Calculating time dilation:
ts
t
v 2 
1  2 
c 
Stationary object
on a moving
spaceship.
t = time (observer)

ts = time if standing still (object)
v = velocity
c = speed of light
Example: The period of a pendulum is measured
to be 3.00 s in the inertial frame of the pendulum.
What is the period measured by an observer
moving at a speed of 0.95c?
ts
t
v 2 
1  2 
c 

3.00
t
.95 2 c 2 
1  2 
 c 
t = 9.6 seconds

If you were moving in a spaceship at a
high speed relative to Earth, would you
notice a difference in your pulse rate?
No, there is no relative speed difference between
you and your pulse.
Would you notice a difference in the
pulse rate of the people left on Earth?
Yes, it would be slower then usual. The
relativistic effect always happens to “the
other guy.”
Does time dilation mean that time really
does pass more slowly in moving systems
or that it only seems to pass more slowly?
• The slowing of time in moving systems
is not merely an illusion resulting from
motion. Time really does pass more
slowly in a moving system compared
with one at relative rest.
• But this leads to some interesting
paradoxes.
The Twin Paradox
There are two twins, Speedo and Goslo. When they are
20 years old, Speedo, the more adventurous of the
two sets off on an epic journey to Planet X, located
20 lightyears away from Earth. Further, his spaceship
is capable of reaching a speed of 0.95c relative to the
inertial frame of his twin brother back home.
After reaching Planet X, Speedo becomes homesick
and immediately returns to Earth at the same speed
of 0.95c. Upon his return, Speedo is shocked to
discover that Goslo is now an old man while Speedo
is not.
The Paradox
From Speedo’s point of view, his brother
Goslo was the one racing away and
then back at 0.95c. Therefore, Goslo
should now be younger then Speedo.
But that isn’t the case.
The Resolution
• Consider a third observer traveling in a spaceship at a
constant speed of 0.50c relative to Goslo. To the third
observer, Goslo never changes inertial reference
frames (his speed relative to the observer is always the
same).
• The third observer notes however, that Speedo
accelerated during his journey, changing his reference
frame in the process.
• To the third observer, the motion of Goslo and Speedo
are not the same. Therefore roles played by Goslo and
Speedo are not symmetric. So it should not be
surprising that time flows differently for each.
• Twin Paradox Video
Example: How old are Goslo and Speedo when they finally
reunite?
(remember, Planet X is 20 lightyears away. The are each 20
years old in the beginning, and Speedo travels at 0.95c)
For Goslo:
Speedo:
2dist. 2(20ly)
t

v
0.95ly/ y
ts
t
v2
1 2
c
t = 42 years
Age = 42 yrs + 20 yrs

Goslo is 62 years old
ts
 42yrs 
0.95 2 c 2
1
c2
ts = 13 years

Age = 13yrs + 20 yrs
Speedo is 33 years old
E=
2
mc
Probably the most famous scientific
equation of all time, first derived by
Einstein is the relationship E = mc2.
This tells us the energy corresponding to a mass, m, at
rest. What this means is that when mass disappears, for
example in a nuclear fission process, this amount of energy
must appear in some other form. It also tells us the total
energy of a particle of mass m sitting at rest.
E=
2
mc
"It followed from the special theory
of relativity that mass and energy are
both but different manifestations of the
same thing -- a somewhat unfamiliar
conception for the average mind.
Furthermore, the equation E is equal to
m c-squared, in which energy is put equal to mass,
multiplied by the square of the velocity of light, showed that
very small amounts of mass may be converted into a very
large amount of energy and vice versa. The mass and
energy were in fact equivalent, according to the formula
mentioned above. This was demonstrated by Cockcroft and
Walton in 1932, experimentally."
What E = mc2 doesn’t mean:
Many people think that E = mc2 means matter,
when traveling at the speed of light
transforms into pure energy.
THIS IS NOT TRUE
1. Matter can’t travel at the speed of light.
2. The speed of light squared is not a velocity.
All E = mc2 means is that mass and energy are
two sides of the same coin. They are the
same thing.
Energy
• Rest energy (also called Rest Mass) - The energy a particle
has by simply being a particle (having mass). Also called
mass-energy
Erest = mc2
• Kinetic Energy - Energy due to the movement of the
particle
2
mc
2
KE 

mc
1 (v 2 /c 2 )
• Total Energy - The total energy a particle has due to
the fact it is a particle (rest energy) and the fact that it
is moving (kinetic energy)

mc
E total  E rest  KE 
2
2
1 (v /c )
2
As a student, Einstein
was a whiz in math &
science, but he was
lousy in grammar and
home economics.
Example: A nickel has a mass of 5.00 g. If this mass could
be converted to electric energy, how long would it keep a
100. W light bulb lit? (remember 1 W = 1 J/sec)
First: How much energy is equivalent to 5.00 g
(physics prefers kg!)
Erest  mc 2
Erest  (0.005kg )(3.0 x108 m / s) 2
Erest  4.5 x1014 J
Second: 100 W = 100. J/sec. We have ______ J,
how many seconds is that?
1sec
12
4.5 x10 J 
 4.5 x10 sec
100 J
or 143,000 years!
14
Fusion & Fission processes take advantage of a
particle’s rest energy (E = mc2) & change mass into
energy.
Fusion processes in stars fuse hydrogen atoms
together to form helium atoms according to the
following (simplified) process:
41H  4He + energy
• If one mole of 1H = 1.007824 g
• & one mole of 4He= 4.002603 g,
A) How much mass is lost in this process?
Dmass = 4.002603 – 4(1.007824) = - 0.028693 g
Or 2.8693 x10-5 kg
B) Where did this mass go/what did it turn in to?
Energy, specifically light & heat
C) How much energy was created when 2.8693x10-5 kg
of mass turned into energy?
E  mc 2
5
E  (2.8693  10 kg)(3.0  10 m/s)
E  2.58237  10 J
12
8
2